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Dec 1, 2020 at 3:24 answer added Bruce Abbott timeline score: 1
Dec 1, 2020 at 2:01 comment added user57037 Base to emitter voltage depends on base current and temperature of the transistor. Hotter temps = lower voltage for same current (this is linear). Higher current = higher voltage, but it is non-linear. Doubling the base current will only make the base voltage slightly higher.
Dec 1, 2020 at 1:56 review Close votes
Dec 18, 2020 at 3:02
Dec 1, 2020 at 1:56 answer added user57037 timeline score: 1
Dec 1, 2020 at 1:42 comment added TimWescott Voting to close on the grounds that this needs more clarity. Specifically, you need to ask just one question at a time -- you're asking a whole bunch of unrelated questions, and your later questions are based on erroneous assumptions about the answers to the earlier questions. Why don't you edit your question to stop at the first question mark. We'll answer that, then you can post the next one, we'll work through that, etc.
Dec 1, 2020 at 1:22 comment added Sredni Vashtar "The voltage drop between collector and emitter is a fixed value for the transistor made of certain material (0.7V for silicon and 0.2V for germanium)" No, it is not fixed and those are the (conventional) values for Vbe. Also, that diagram is misleading to say the least. The operating point of the transistor is constrained to be on the intersection of the load line with one of the output curves. It will not fall in the shaded areas.
Dec 1, 2020 at 1:18 comment added jonk Don't you want the BJT to stay in the active region? (I've a hunch, reading through your questions, that you have many thoughts that need to be tossed overboard and then new ones acquired.)
Dec 1, 2020 at 1:03 history edited P L CC BY-SA 4.0
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Dec 1, 2020 at 0:56 review First posts
Dec 8, 2020 at 9:48
Dec 1, 2020 at 0:55 history asked P L CC BY-SA 4.0